# Quotient of ideal in group ring is isomorphic to abelianization [duplicate]

Let $G$ be a group and $\mathbb Z G$ the group ring over the integers. Let $I$ be the ideal of elements $\sum_{g\in G} n_g g$ with $\sum_{g\in G} n_g = 0$. I am trying to prove that $I/I^2$ is isomorphic (as an additive abelian group) to the abelianization $G/[G,G]$. I have tried showing that it satisfies the universal property of the abelianization, but this does not seem to work. I also tried constructing an explicit isomorphism, but could not find one. Any ideas?

## marked as duplicate by user26857, Lord_Farin, Brandon Carter, Micah, rschwiebMay 8 '13 at 16:25

We have here the augmentation ideal $\,I\,$, generated by $\,\{x-1_G\;\;;\;x\in G\}\,$ . Define now
$$\phi: G\to I/I^2\;\;,\;\;\phi(x):=x-1_G+I^2$$
(1) Prove $\,\phi\,$ is a homomorphism of groups
(2) Deduce that $\,G'\leq\ker\phi\,$
(3)Prove that $\,\psi:I\to G/G'\,$ defined by $$\psi\left(\sum_{x\in G}n_x(x-1_G)\right):=\prod_{x\in G}x^{n_x}G'\,$$ is an inverse homomorphism to $\,\phi\,$ ...